PS2Solutions - 1 APSC 132 PROBLEM SET 2 SOLUTIONS Problem 1...

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1 APSC 132 PROBLEM SET 2 SOLUTIONS Problem 1 The expansion is adiabatic and irreversible . P gas V gas V 1 = V 2 = n = 1.00 mol n = 1.00 mol P 1 = 5050 kPa P 2 = 101 kPa T 1 = 773 K T 2 = C v = C p – R = 21.0 J/mol.K V 1 = nRT 1 /P 1 = (1.00)(8.31)(773)/5050 L = 1.27 L Because process is irreversible , we cannot use P 1 V 1 g =P 2 V 2 g . However, we still require T 2 . Because process adiabatic, D E = w (because q = 0). Therefore, nC v (T 2 -T 1 ) = -P ext (V 2 -V 1 ) = -P ext (nRT 2 /P 2 – nRT 1 /P 1 ) =-nRP ext (T 2 /P 2 – T 1 /P 1 ) Unless stated otherwise, you can assume that P ext = P 2 . We know all variable except T 2 . Solve for T 2 . (1.00)(21.0)(T 2 – 773) = -(1.00)(8.31)(101)(T 2 /101 – 773/5050)
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2 T 2 = 558 K, D T = 558 K – 773 K = -215 K w = D E = nC v D T = (1.00)(21.0)(-215) J = -4515 J D H = nC p D T = (1.00)(29.3)(-215) J = -6300 J q = 0 J Problem 2 a) 2 isothermal expansion P gas adiabatic compression isobaric compression 1 3 V gas b) What do we know? P
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This note was uploaded on 12/01/2010 for the course APPLIED SC 131 taught by Professor Professornewstead during the Fall '09 term at Queens University.

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PS2Solutions - 1 APSC 132 PROBLEM SET 2 SOLUTIONS Problem 1...

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